# Calculate the amount of aluminum required to get 1120Kg of iron by the reaction. 2Al + Fe2O3 --> Al2O3 + 2Fe?

2
by ob6katFujegu

2016-03-16T23:27:44+05:30

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According to the balanced equation.
2 moles of Al(54kg) gives 2 moles of Fe(112kg).
So for 1120 kg of Fe we need  540 kg of Al.
2016-03-17T15:53:17+05:30

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As per the equation
2 mol of Aluminium + 1 mole of Iron Oxide -----> 1 mole of Aluminium Oxide + 2 moles of Iron

1 mole of aluminium = 27 U
1 mole of Iron Oxide =  56*2 +16*3 = 160U

1 mole of Aluminium Oxide = 27*2 + 16*3 = 102U

54U  +  160U  →  102U+112U 54  g  +  160g  →  102g+112g As  per  the  balanced  equation  54g For  112g  of  iron,  we  required  54g  Aluminum  for  1120  kg  of  iron  required  aluminum  is-

(1120*1000) *54 =540000 gram
so it is 540 kg